Question

find the volume of the solid generated by the following equations when rotated around the given axis: y=sqrtx y=0 and x=3
a)the y axis
b)the line x=3
c)the line x=6

Answer 1

yo amistre i need some help on some integrals as well... idk why i cant get them right

Answer 2

we can shell it from 0 to 3 to rotate it about the x=0 axis

Answer 3

cant shell, only washer and disk

Answer 4

why cant you shell?

Answer 5

not allowed to haha im not supposed to know how to shell yet

Answer 6

do you gotta show your work? or just get the answer?

Answer 7

well the thing is is that i can get the answer from the back, its the work that i cant get right for some reason

Answer 8

to disk it around the x=0 or x=3 axis; we solve the eqs for y and intf(y)dy it

Answer 9

are the limits also in terms of y? so 0 to sqrt3 ?

Answer 10

yes; since you are traveling along the y axis to add up all the parts

Answer 11

but i stillg et the wrong answer O.o

Answer 12

y = 0 to y = sqrt(3)

Answer 13

first figure the volume for the "cylinder" that is created by a radius of 3 and a height of sqrt(3)

Answer 14

i use:
\[\Pi \int\limits_{0}^{\sqrt{3}} (y ^{2})^{2}dy\]

Answer 15

that is the area you wanna subtract yes ?

Answer 16

subtract?

Answer 17

i thought thats just the area i wanted XD

Answer 18

oh i see. so i want to subtract it from
the cylinder which is r=3 h=sqrt3?

Answer 19

subtract the area created by the lower function from the area created by the upper function

Answer 20

yes

Answer 21

your making a funny looking donut

Answer 22

wait am i finding the area of the shaded, or the white?

Answer 23

the shaded... right?

Answer 24

the shaded is the area of the boundary; now we got to spin it about the x=0 axis; the x=6 axis and the x=3 axis for each part of the question

Answer 25

the shaded area tells you what is spinning; rotate it to find the volume of revolution :)

Answer 26

iight so for the x=3 and x=6... i cant get my euation straight ...

Answer 27

why not: just subtract 3 and 6 respectively from the equations

Answer 28

why subtract those?

Answer 29

why not add or such?

Answer 30

you want to get your axis of rotation to the origin its easier to calculate from that point

Answer 31

sqrt(x-3) spins about the origin; sqrt(x-6) move the x = 6 axis to the origin etc..

Answer 32

the heights never change :) do they

Answer 33

ok i get it, idk why im getting so lost with concepts i used to know cold -.-

Answer 34

keep in practice :)

Answer 35

can you show me how to evaluate the integral of sech(x/2)

Answer 36

need more practice.. If you get rusty, just find someone to tutor. That's what I do. ;)

Answer 37

sech? cant say that Ive had to much practice with hyperbolics ...

Answer 38

i come up with 4arctan(x/2) but then apparently because i have the area bounded by -4 and 4, it makes it 8arctan(x/2)-2pi... WHY!?!?!??!

Answer 39

well im sure if you treat it as 2/(e^x + e^-x) you can do it XD

Answer 40

wolframalpha it and see what they can shed some light on

Answer 41

in this case, x is (x/2)

Answer 42

wolframalpha didnt help understand why it did that tho

Answer 43

polpak.. can you do it? XD

Answer 44

sech is an even function. So \[\int_{-a}^a f(x)dx = 2\int_0^af(x)dx\]

For the even function you come up with for the rotation.

Answer 45

.... oh wow, what about the -2pi ?

Answer 46

and for the solids of rotation, when done around the x=3 and x=6, do my bounds change? how?

Answer 47

Sorry, someone called. just a sec.

Answer 48

yeahh buddy

Answer 49

Changing the bounds on x will only change the limits of your integration if you are still rotating about the x axis. If you are rotating about some other line, then things get tricky.

Answer 50

explainnn please :O

Answer 51

explain what?

Answer 52

how the bounds change

Answer 53

The best thing to use here is the cylindrical coordinates, since you need the x, y axes to find the area on 2-d and then sweep it across the axis of your interest to find the volume. Follow my next post to find the procedure of computing the area when rotated across y-axis (question a), the others can be similarly found. The answer is
\[4\sqrt{3} \pi\]

Answer 54

\[\int\limits_0^{2\pi} \int\limits_0^3 \int\limits_0^{\sqrt{x}} dy dx d \theta\]
This is the volume when you rotate around y-axis. Carry out the calculations, I guess I did the calculations correct, the answer is as above I mentioned. Here is a drawing of the procedure.